Interpolating Between Hyperbolic and Sol Geometry

A smooth one-parameter family connecting two Thurston geometries.

This short note gives a metric on $\mathbb{R}^3$ which in some sense interpolates between 3-dimensional hyperbolic space and Sol geometry.

The Log Model of the Hyperbolic Plane

The first step is to build a new model of the hyperbolic plane on all of $\RR^3$, by taking the logarithm of the $z$ coordinate. Precisely, let $L,E$ be the pair of homeomorphisms $\mathcal{H}\leftrightarrow \RR^3$ given by $L(x,y,z)=(x,y,\log z)$, $E(x,y,z)=(x,y,e^z)$. We have the hyperbolic metric written on $\mathcal{H}$, so we can use $E$ to pull it back to $\RR^3$.

PICTURE

To make the calculation clear, we will temporarily use $X,Y,Z$ as the coordinates on $\mathcal{H}$ and $x,y,z$ the coordinates on $\RR^3$. The metric on $\RR^3$ is determined by all the pairwise dot-products of the coordinate basis fields $\partial_x,\partial_y,\partial_z$, which can be computed via pullback after understanding the action of $E$ at a point $(x,y,z)$:

$E_\ast \partial_x = \partial_X\hspace{1cm}E_\ast \partial_y =\partial_Y\hspace{1cm}E_\ast\partial_z=e^z\partial_Z$

For mixed dot products, note that $E$ does not change the angles between basis directions so $\partial_x,\partial_y,\partial_z$ are pairwise orthogonal under pullback just as $\partial_X,\partial_Y,\partial_Z$ are on $\mathcal{H}$. This leaves only the diagonal terms to compute:

$$\begin{align} \langle \partial_x,\partial_x\rangle_{(x,y,z)}&=\langle E_\ast \partial_x,E_\ast \partial_x\rangle_{E(x,y,z)}\\ &= \langle \partial_X,\partial_X\rangle_{(x,y,e^z)}\\ &=\frac{1}{(e^z)^2}\left(\partial_X\cdot\partial_X\right)\\ &=\frac{1}{e^{2z}} \end{align}$$

Where in the last line we used the metric on $\mathcal{H}$ is the Euclidean metric divided by the square of the $z$ coordinate. The same reasoning holds for $\langle \partial_y,\partial_y\rangle$ giving an identical answer. This leaves only the $z$ component:

$$\begin{align} \langle \partial_z,\partial_z\rangle_{(x,y,z)}&=\langle E_\ast \partial_z,E_\ast \partial_z\rangle_{E(x,y,z)}\\ &= \langle e^z\partial_Z,e^z\partial_z\rangle_{(x,y,e^z)}\\ &=\frac{1}{(e^z)^2} \left(e^z\partial_Z\cdot e^z\partial_z\right)\\ &=\frac{(e^z)^2}{(e^z)^2}(\partial_Z\cdot \partial_Z)\\ &= 1 \end{align}$$

Thus the metric tensor in coordinates $(x,y,z)$ on $\RR^3$ is

$g = e^{-2z}dx^2+e^{-2z}dy^2+dz^2$

This metric has the factors in front of the $z$ coordinate exponentially decreasing as $z$ increases: it will be more convenient to ‘flip’ this behavior, and pull back the metric once more under the reflection $(x,y,z)\mapsto(x,y,-z)$. The result is simply to flip the sign of the exponents:

The Interpolating Metric

The log-hyperbolic model looks very similar to Sol geometry: they have the same coordinate domain and the metric form is identical save a crucial difference: in $\mathbb{H}^3$ the sign of the exponentials is both the same, and in Sol the exponential factors prefixing the $x$ and $y$ coordinate directions are opposite. A natural idea is to change the prefactor on $dx^2$ to a different function of $z$,

$g= f(z)dx^2+e^{2z}dy^2+dz^2$

Where $f(z)\approx e^{2z}$ for $z>>0$ and $f(z)\approx e^{-2z}$ for $z << 0$. This would produce a geometry which looks much like $\mathbb{H}^3$ for $z>>0$ and much like Sol for $z << 0$. The obvious choice of such a function is some sort of hyperbolic cosine, as this is built directly from these two exponentials. Writing things down explicitly, one sees that $2\cosh(2z)$ is the correct form:

$$\begin{align} g &= 2\cosh(2z) dx^2+e^{2z}dy^2+dz^2\\ &= \left(e^{2z}+e^{-2z}\right)dx^2+e^{2z}dy^2+dz^2 \end{align}$$

More work needs to be done to understand this geometry: in particular, it remains to be quantified exactly how similar the geometries on each side of $z=0$ approach hyperbolic and sol.